\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Path Integrals

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Path Integrals

Let's start off with a video. This is a great video which includes a complete introduction, with examples, and explanation of applications of path integrals. It is well worth your time to watch it.

Dr Chris Tisdell - Integration over curves [45mins-37secs]

video by Dr Chris Tisdell

The equations from the last video are important for evaluating path integrals. Here is a summary.

requirements

 

result

continuous scalar function \(f(x,y,z)\)

 

path integral of \(f\) over \( \mathcal{C} \)

curve \( \mathcal{C} \) parameterized as \(\vec{c}(t)\) on an interval \(a \leq t \leq b\)

 

\(\displaystyle{ \int_{\mathcal{C}}{ f(x,y,z)~ds} = }\) \(\displaystyle{ \int_{a}^{b}{ f(\vec{c}(t)) ~ \| \vec{c}'(t) \| ~dt } }\)

continuous \(\vec{c}'(t)\) [derivative of \(\vec{c}(t)\) with respect to \(t\)]

 

Most of the above equations should be familiar to you. However, one comment is in order about the term \(f(\vec{c}(t))\). How do you substitute a vector into a function?
If we write the vector as \( \vec{c}(t) = \langle X(t), Y(t), Z(t) \rangle = X(t) \hat{i} + Y(t) \hat{j} + Z(t)\hat{k} \) then we substitute the vector components into \(f(x,y,z)\) as \(x=X(t)\), \(y=Y(t)\) and \(z=Z(t)\). Another way to write this is \(f(x,y,z) = f(X,Y,Z) = f(X(t), Y(t), Z(t))\). The result is a function of \(t\) only. All \(x\)'s, \(y\)'s and \(z\)'s will be gone, leaving only an integral in \(t\).

A very interesting comment he makes in the above video is that the direction of the curve does not affect the answer, i.e. integrating in the opposite direction will give the same answer. This is true because we use only the magnitude of the derivative of the parameterized curve, i.e. \(\| \vec{c}'(t) \|\), which is the same regardless of the direction. However, the path integral is dependent on the chosen path, i.e. different paths will, in general, give different results. So path integrals are sensitive to the paths.

Okay, let's try some practice problems.

Practice

Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.

Problem Statement

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}} x+y ~ds \) over the curve \( \mathcal{C}: \) straight line from \((0,0)\) to \((1,1)\), giving your answer in exact terms.

Solution

858 video

video by Dr Chris Tisdell

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.

Problem Statement

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 2\pi \), giving your answer in exact terms.

Solution

859 video

video by Dr Chris Tisdell

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.

Problem Statement

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{x+y+z ~ ds } \) over the curve \( \mathcal{C}: \) straight line \( (1,0,0) \) to \( ( 1,0, 2\pi) \), giving your answer in exact terms.

Solution

860 video

video by Dr Chris Tisdell

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Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^2-1+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.

Problem Statement

Unless otherwise instructed, evaluate the path integral \( f(x,y,z) = x^2+y^2-1+z \) over the curve \( \mathcal{C}: \langle \cos(t), \sin(t), t \rangle; 0 \leq t \leq 3\pi \), giving your answer in exact terms.

Solution

854 video

video by Dr Chris Tisdell

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\) - \((1,1)\) and \((1,1)\) - \((1,0)\), giving your answer in exact terms.

Problem Statement

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{ x+y^2 ~ds }\) over the curve \( \mathcal{C}: \) 2 lines \((0,0)\) - \((1,1)\) and \((1,1)\) - \((1,0)\), giving your answer in exact terms.

Solution

861 video

video by Dr Chris Tisdell

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Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.

Problem Statement

Unless otherwise instructed, evaluate the path integral \( \int_{\mathcal{C}}{xy^4 ~ds } \) over the curve \( \mathcal{C}: \) right half of the circle \( x^2+y^2 = 16 \), giving your answer in exact terms.

Solution

862 video

video by PatrickJMT

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Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).

Problem Statement

Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).

Final Answer

\((5\sqrt{5}+11)/6\)

Problem Statement

Evaluate \( \int_{\mathcal{C}}{ 2x ~ds } \) where \(\mathcal{C} = \mathcal{C_1} + \mathcal{C_2}\); \(\mathcal{C_1}\) goes from \((0,0)\) to \((1,1)\) along \(y=x^2\); \(\mathcal{C_2}\) goes from \((1,1)\) to \((1,2)\) along \(x=1\).

Solution

2226 video

video by Michel vanBiezen

Final Answer

\((5\sqrt{5}+11)/6\)

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Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).

Problem Statement

Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).

Hint

There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1-t)x_1+tx_2\)
\(y=(1-t)y_1+ty_2\)
\(z=(1-t)z_1+tz_2\)
\(0 \leq t \leq 1\).

Problem Statement

Evaluate \(\int_{\mathcal{C}}{x^2z~ds}\), where \(\mathcal{C}\) is the line segment from \((0,6,2)\) to \((-1,8,7)\).

Hint

There are an infinite number of ways to parameterize this curve. This instructor uses the equations
\(x=(1-t)x_1+tx_2\)
\(y=(1-t)y_1+ty_2\)
\(z=(1-t)z_1+tz_2\)
\(0 \leq t \leq 1\).

Solution

2298 video

video by MIP4U

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

related 17calculus pages

line integrals

external links you may find helpful

Pauls Online Notes - Line Integrals Part 1

Wikipedia - Line Integral

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, evaluate these path integral over the curve \(\mathcal{C}\), giving your answers in exact terms.

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